Question

point z is the incenter of δsrt. what is m∠ztb? 24° 31° 35° 62°

Explanation:

Step1: Recall incenter properties

The incenter \( Z \) of a triangle is the intersection of angle bisectors. So, \( ZB \perp ST \) (since \( B \) is the foot of the perpendicular from \( Z \) to \( ST \)), \( ZA \perp SR \), and \( ZC \perp RT \). Also, we can find \( \angle SRT \) and \( \angle RST \) first.

From the diagram, \( \angle RST = 24^\circ + 24^\circ = 48^\circ \), \( \angle SRT = 35^\circ + 35^\circ = 70^\circ \).

Step2: Find \( \angle STR \)

In \( \triangle SRT \), the sum of angles is \( 180^\circ \). So, \( \angle STR = 180^\circ - \angle RST - \angle SRT \).
Substitute values: \( \angle STR = 180^\circ - 48^\circ - 70^\circ = 62^\circ \).

Step3: Use angle bisector for \( \angle STR \)

Since \( Z \) is the incenter, \( TZ \) bisects \( \angle STR \). So, \( \angle ZTB = \frac{1}{2} \angle STR \).
Substitute \( \angle STR = 62^\circ \): \( \angle ZTB = \frac{62^\circ}{2} = 31^\circ \).

Answer:

\( 31^\circ \) (corresponding to the option "31°")